15 THE LEBESGUE-NIKODYM THEOREM 193 are locally /l-integrable real-valued functions. Hence the assertions of (i) are consequences of those of (ii). To prove (126.96.36.199), we reduce to the case where n2 = Q; for if supO*! - yU2 , 0) exists and is equal to (gt - g2)+ - ^ it follows immediately (13.3) that we shall have A*2 > 0) = (g2 + (gl -~ #2)+) • A = sup(^, g2) - L This reduces us to proving (188.8.131.52). For this, we shall begin by proving the last assertion of (ii), or equivalently that the relation^ * A ^ 0 implies g(x) ^ 0 almost everywhere with respect to L Let N be the set of points x e X such that g(x) < 0. If we put v = $~A, it follows from (184.108.40.206) that v(X - N) =0; so if we can show that v(N) s= 0, it will follow that v = 0, and this will estab- lish our assertion (13.14.4). Clearly it is enough to prove that v(N n K) =0 for each compact subset K of X. If U is any relatively compact open set containing N n K, then we have j g~ dl <S f g+ dL For by hypothesis /• J u J u J(#+ -g~)fdk ^ 0 for every function/^ 0 belonging to «2fR(X), and it is therefore sufficient to remark that | g~ dX =sup f g""fdk, ( g+ dl = /« J U J J U supl g+fdA, the supremum being taken over all functions /e JfR(X) such that 0 g/^ ^,j (13.5.1). Since #+(x) = 0 for x e N n K, we have I ^+^ = inf f ^+^-0; JNnK U Ju consequently, f g dk = inf f' g dk= 0, which shows that v(N n K) = 0 JNnK y J U (13.14.3). Now let p be a measure such that p ^ 0 and p ^ g • A. Putting A + p = o-, we have (13.15.1) p = w • a and A = P • <r, where w and y are locally cr-integrable and u g 0 and t? ^ 0 almost everywhere with respect to cr. It follows that u^gv almost everywhere with respect to <j, hence that u^(gv)* =g*v almost everywhere with respect to <r; but this implies that p = u - cr ^ (^+f) * a = #* -A by virtue of (13.14.5). Finally, we have to prove (iii). If the sequence (gn - X) is bounded above in MR(X), then for each function/^ 0 belonging to ^R(X) we have sup \fgndX< +00; n Jlies that v*(/z) ^ /z(/z) ^ e, and the proof is complete.